- A$1$
- ✓$2$
- C$0.5$
- D$2.5$
In parallel, ${I_2} = \frac{E}{{2 + \frac{r}{2}}} = \frac{{2E}}{{4 + r}}$
Since ${i_1} = {i_2}$ $ \Rightarrow $ $\frac{{2E}}{{4 + r}} = \frac{{2E}}{{2 + 2r}}$ $ \Rightarrow $ $r = 2\,\Omega $
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Statement $I:$ If the Brewster's angle for the light propagating from air to glass is $\theta_{ B }$, then Brewster's angle for the light propagating from glass to air is $\frac{\pi}{2}-\theta_B$.
Statement $II:$ The Brewster's angle for the light propagating from glass to air is $\tan ^{-1}\left(\mu_{ g }\right)$ where $\mu_{ g }$ is the refractive index of glass.
In the light of the above statements, choose the correct answer from the options given below :